Moriah 2001 Polynomials Question (1 Viewer)

Sy123 · Nov 11, 2012

Thank You

Riproot · Nov 11, 2012

Wow. What?

seanieg89 · Nov 11, 2012

oh well · Nov 11, 2012

Maffs freaks

Sy123 · Nov 11, 2012

seanieg89 said:
$We have that the derivative of a polynomial which has a root of multiplicity $n>1$ at $x=\alpha$ has a root of multiplicity $n-1$ at $\alpha$. Applying this at $0$ and $1$ (observing that $\frac{d}{dx}(\phi(x)-1)=\phi'(x)$) gives us:\\ $\phi'(x)=Ax^2(x-1)^2.$ for some unknown nonzero constant A. Integrating both sides, we get:\\ $\phi(x)=A(x^5/5-x^4/2+x^3/3)+B$ for some constant $B$. But the divisibility condition implies that $\phi(0)=0,\phi(1)=1.$\\These conditions force $A=30,B=0$. \\Whence $\phi(x)=6x^5-15x^4+10x^3.$$

Ah, I see there. Nice solution, thank you.

seanieg89 · Nov 11, 2012

Sy123 said:
Ah, I see there. Nice solution, thank you.

No worries, I feel like I did something slicker last time but whatever...its late and this works.

Riproot · Nov 11, 2012

seanieg89 said:
$We have that the derivative of a polynomial which has a root of multiplicity $n>1$ at $x=\alpha$ has a root of multiplicity $n-1$ at $\alpha$. Applying this at $0$ and $1$ (observing that $\frac{d}{dx}(\phi(x)-1)=\phi'(x)$) gives us:\\ $\phi'(x)=Ax^2(x-1)^2.$ for some unknown nonzero constant A. Integrating both sides, we get:\\ $\phi(x)=A(x^5/5-x^4/2+x^3/3)+B$ for some constant $B$. But the divisibility condition implies that $\phi(0)=0,\phi(1)=1.$\\These conditions force $A=30,B=0$. \\Whence $\phi(x)=6x^5-15x^4+10x^3.$$

I feel so stupid now.

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Moriah 2001 Polynomials Question (1 Viewer)

Sy123

This too shall pass

Riproot

Addiction Psychiatrist

seanieg89

Well-Known Member

oh well

Banned

Sy123

This too shall pass

seanieg89

Well-Known Member

Riproot

Addiction Psychiatrist

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